Let y=y(x) solve (y^2-x)dydx=1, y(0)=1.

Question

Let $y=y(x)$ solve $(y^2-x)\dfrac{dy}{dx}=1$, $y(0)=1$. This curve intersects the $x$-axis at abscissa:

Accepted Answer

$\dfrac{dx}{dy}+x=y^2$ (linear in $x$). I.F.$=e^y$.
$$xe^y = \int y^2e^y\,dy = e^y(y^2-2y+2)+C.$$

Using $x=0,y=1$: $0=e(1)+C \implies C=-e$.

At $y=0$: $x=2+(-e) = 2-e$.

Solutions · 1